1. **Problem statement:** We have a rectangular sheet of metal 20 cm by 12 cm. Squares of side length $x$ cm are cut from each corner, and the sides are bent up to form an open box. We need to find the value of $x$ that maximizes the volume of the box and determine that maximum volume. 2. **Formula for volume:** After cutting squares of side $x$, the new dimensions of the box base will be $(20 - 2x)$ cm by $(12 - 2x)$ cm, and the height will be $x$ cm. The volume $V$ of the box is given by: $$V = x(20 - 2x)(12 - 2x)$$ 3. **Expand the volume expression:** $$V = x[(20)(12) - 2x(20) - 2x(12) + 4x^2] = x[240 - 40x - 24x + 4x^2] = x[240 - 64x + 4x^2]$$ Multiply $x$ inside: $$V = 240x - 64x^2 + 4x^3$$ 4. **Find critical points by differentiation:** Take the derivative of $V$ with respect to $x$: $$\frac{dV}{dx} = 240 - 128x + 12x^2$$ Set derivative equal to zero to find critical points: $$240 - 128x + 12x^2 = 0$$ Divide entire equation by 4 for simplicity: $$60 - 32x + 3x^2 = 0$$ Rewrite: $$3x^2 - 32x + 60 = 0$$ 5. **Solve quadratic equation:** Use quadratic formula: $$x = \frac{32 \pm \sqrt{(-32)^2 - 4 \cdot 3 \cdot 60}}{2 \cdot 3} = \frac{32 \pm \sqrt{1024 - 720}}{6} = \frac{32 \pm \sqrt{304}}{6}$$ Calculate $\sqrt{304} \approx 17.4356$: $$x_1 = \frac{32 + 17.4356}{6} \approx 8.57$$ $$x_2 = \frac{32 - 17.4356}{6} \approx 2.43$$ 6. **Check domain constraints:** Since squares are cut from corners, $x$ must be less than half the smaller side: $$x < \frac{12}{2} = 6$$ So $x_1 \approx 8.57$ is invalid. 7. **Evaluate volume at $x = 2.43$:** $$V = 2.43(20 - 2 \cdot 2.43)(12 - 2 \cdot 2.43) = 2.43(20 - 4.86)(12 - 4.86) = 2.43 \times 15.14 \times 7.14 \approx 262.8$$ 8. **Check volume at endpoints:** At $x=0$, $V=0$. At $x=6$, $V=0$. So the maximum volume is approximately $262.8$ cubic centimeters at $x \approx 2.43$ cm. **Final answer:** The maximum volume of the box is approximately **262.8 cm³** when squares of side length **2.43 cm** are cut from each corner.